Problem: Simplify and expand the following expression: $ \dfrac{1}{t - 1}+ \dfrac{4}{2t - 20}- \dfrac{3}{t^2 - 11t + 10} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the second term: $ \dfrac{4}{2t - 20} = \dfrac{4}{2(t - 10)}$ We can factor the quadratic in the third term: $ \dfrac{3}{t^2 - 11t + 10} = \dfrac{3}{(t - 1)(t - 10)}$ Now we have: $ \dfrac{1}{t - 1}+ \dfrac{4}{2(t - 10)}- \dfrac{3}{(t - 1)(t - 10)} $ The least common multiple of the denominators is: $ (t - 1)(t - 10)$ In order to get the first term over $(t - 1)(t - 10)$ , multiply by $\dfrac{2(t - 10)}{2(t - 10)}$ $ \dfrac{1}{t - 1} \times \dfrac{2(t - 10)}{2(t - 10)} = \dfrac{2(t - 10)}{(t - 1)(t - 10)} $ In order to get the second term over $(t - 1)(t - 10)$ , multiply by $\dfrac{t - 1}{t - 1}$ $ \dfrac{4}{2(t - 10)} \times \dfrac{t - 1}{t - 1} = \dfrac{4(t - 1)}{(t - 1)(t - 10)} $ In order to get the third term over $(t - 1)(t - 10)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{3}{(t - 1)(t - 10)} \times \dfrac{2}{2} = \dfrac{6}{(t - 1)(t - 10)} $ Now we have: $ \dfrac{2(t - 10)}{(t - 1)(t - 10)} + \dfrac{4(t - 1)}{(t - 1)(t - 10)} - \dfrac{6}{(t - 1)(t - 10)} $ $ = \dfrac{ 2(t - 10) + 4(t - 1) - 6} {(t - 1)(t - 10)} $ Expand: $ = \dfrac{2t - 20 + 4t - 4 - 6}{2t^2 - 22t + 20} $ $ = \dfrac{6t - 30}{2t^2 - 22t + 20}$ Simplify: $ = \dfrac{3t - 15}{t^2 - 11t + 10}$